>> Welcome back. We'll continue with our development of, our one dimensional elliptic partial differential equation and we have been doing it so far using elasticity as a canonical physical problem, right, to guide us through this process. So we had got as far as writing out the differential equation and we were talking about boundary conditions. So let's continue talking about, about boundary conditions, okay? So let me actually draw our domain once again. We have the bar. We have our coordinate x. We have x equals 0 and we have x equals L. Now, At the left end, x equals 0. We have u 0 equals u naught. Right? Specified as a condition. And at the right end, we have either u at L equals u sub g for given, or we talked about how we have sigma at L equals t, okay? And at the very end of the previous segment, we had talked about you know, the nomenclature that we employ for these different boundary conditions. Now I want to point out something that's actually quite important both from a mathematical and a physical standpoint for this problem. Which is the following, observe that we always have this the Dirichlet boundary condition at x equals 0. However, at x equals L, we have either a Dirichlet boundary condition or a Neumann boundary condition. Okay? In, in particular we do not have a situation where at both ends we may have Neumann boundary conditions. Okay. So let me state that. Okay. So in terms of boundary conditions >> Okay? We have either >> case a, being Dirichlet boundary conditions. As we go on, I'll be offering you as BCS short for boundary conditions. So either way, we have Dirichlet boundary conditions at x equals 0 and x equals L, right? Or b. Dirichlet boundary condition at x equals 0 and Neumann boundary condition at x equals L. It turns out that in case b we, we could actually of flipped things around. Which is that we could have had Dirichlet boundary condition x equals l, and a Neumann boundary condition at x equals zero that would not pose a problem. Okay, what's important is to note that there is a particular combination that we are not considering. Okay? It's very obvious but I'll let you think about it for a second or two, to make sure that you are indeed with me. This. Okay. The condition we are not considering is Neumann at both ends. Okay? So, we do not consider Neumann boundary conditions at x equals zero and x equals L. All right? There's a very good reason for this. Okay? Let me try and motivate the reason first from a, from a physical standpoint, okay? What will, why this is important is the following, is, is, is the following. For the kind of problem we are considering, we're saying that we always need to have the displacement specified in one end or the other, at least at one end, okay? So we may have the dis, the displacement specified at this end of where I'm holding it and on, on the other end we may either specify the displacement or a force. That's okay. Right? Or attraction, sorry. That's okay. Okay? Or what we are not allowed to do is to say that we have the traction specified at one end as well at, as, as at the other. Okay? So we can't have Neumann boundary conditions on both sides, okay? And now you may wonder why this is the case. In particular you may ask the question, but surely that leaves out a very important class of problems. In particular, that does leave, it seems to leave out the class of problems where I mean I'll take this bar, right? Remember it's after all an, an elasticity problem, right. You may say that it, we're leaving out the pro, the class of problems where we take this bar and sort of throw it one way. Remember we're doing it one way, so we're assuming that it moves only one direction along the x direction. If we, if we, if we toss this bar and it moves along the x direction, we may say well, we've ruled out the ability to solve that problem because we have stated, so far in our development, that we always have the displacement specified at one end or the other. Okay? If we were to toss it and just watch it, watch it evolve in time, we would not be specifying the displacement at one end, right? It would just be, we would be throwing it and let it evolve in, in space and time, but we would not have specified the displacement at one end. Okay? The reason is that, the reason we, we, we have this restriction is because we are considering steady state problems. The problem where we actually throw the bar and watch it evolve in, in, in space and time would be a time dependent problem. Okay? And in particular in the context of elasticity it would no longer be an elliptic problem, it would be a hyperbolic problem. There are other ways in which that problem is treated. In particular, we would need to specify an additional set of conditions. Can you think of what those may be? We would need initial conditions as well. Okay? What does that mean mathematically, 'kay? What it means mathematically is the following, okay? The reason we do not consider Neuman boundary conditions in, in this description of the problem is because of the following, okay? Supposing you consider, let, let's suppose we do try to specify non boundary conditions at both ends. Okay? So consider, Neumann boundary conditions at x equals 0 and x equals L, okay? So what this may mean is that we are saying that sigma at 0 which we know to be E u,x evaluated at x equals 0 equals let's say. T, 0 right for traction at zero? And sigma at L which is what we have indeed considered is E times u, x evaluated at x equals L now equal to T sub L. Okay? I'm distinguishing T0, T not and T L just by the positions, okay? Now you observe that this poses a certain problem because supposing so, so if you had these two conditions Our, problem would be one of finding this basement field which, in addition to satisfying these two boundary conditions also, of course, needs to satisfy the differential equation. Right? So we had this and now we say consider U of X satisfying these Neumann boundary conditions and the differential equation. Right and the differential equation remember is the sigma d x plus f equals 0. Right? In 0, L. Okay. Now, if you had this sort of situation, making our substitution come from the constitutive relation tells us that now d dx of E times u,x, where I've simply substituted our constitutive relation for this stress, plus f equals 0, okay? Now. This is the problem we would be solving, right? We, we would be solving a problem with these two boundary conditions and this, differential equation. Now observe that any solution we get to this problem would be non unique up to a constant displacement field. Okay? Because supposing you did have a u that satisfied these conditions. Right? Alright. What we could do is that u of x plus some other u which I will denote as u bar. Okay? Which is a constant. With respect to wrt being short for with respect to. X, okay? This field is also a solution. Right? Clearly if u, if u bar is a constant with respect to x, it would satisfy our two Neumann boundary conditions. It would satisfy that. And it would satisfy that because derivatives with respect to u, sorry, derivatives with respect to x of u bar would be zero. It would not contribute to that boundary condition. Likewise to the differential equation, right? Because the differential equation actually involves two derivatives on the displacement field. Right? If, if, if E were a constant. Or even if E were not a constant it would involve at least one derivative on, on, on the displacement field. Right? So therefore adding on U bar equals constant would still satisfy that, that equation as well. Okay? So what we see is that If we had Neumann the boundary conditions only on our problem the solution u of x is non unique. [SOUND] Up to a constant displacement field and in the context of elasticity, a constant displacement field is also called a rigid body motion. Right? Because remember, this is our one dimensional domain over which we're trying to solve this problem. What is a displacement field that is constant with respect to position? It's one in which we seeing, we're seeing that the entire bar has a translation, right? Every point in the bar has the same displacement, u bar, added on to any solution we specify to it. Okay? So we say that the solution u x is non-unique up to a rigid body motion. U bar, okay, which is a constant. Okay? In this particular say, case, saying a rigid body motion and saying u bar is a constant are actually superfluous. They both mean the same thing. Okay? So it is that for these kinds of problems that we are looking at, in particular elliptic differential equations in one dimension, okay? We cannot have Neumann boundary conditions, only. We do need a traditionally boundary condition and it is exactly to make sure that our solution is unique with respect to the solution, with respect to the primary field. Okay? And also remember that I said that, that, we did talk about how the problem of actually tossing this bar and watching it move in space and time is a different problem. Okay it's no longer an elliptic problem. In particular, it's no longer a steady state problem. Because we would be looking at how a solution evolves in time by not assuming that things are constant In time. Okay. So maybe I should state that as well. Neumann boundary conditions alone can be specified. For the time-dependent. Elasticity problem. Okay? And I'll just state here that though we won't be coming back to this time-dependent elasticity problem. I will just state for now that this would then become what is called a hyperbolic differential equation. The hyperbolic partial differential equation. Okay? We are not doing that right now. We are looking at elliptic partial differential equations, which for the kinds of things we are looking at also means that we have steady state problems, okay? So those are the important things we need to say about the boundary conditions. What I will also do now is spend a few minutes talking about what I've been calling our body force, okay? So let us now recall. The body force that we've written as f has a function of x. Okay, and the role it plays in our problem is the following. It gives us d sigma dx plus f, and here I'll make it explicit that it depends upon position, is equal to 0 in. Okay? In this case, what I'm calling the body force, of course is a term that is applicable only when we're doing elasticity. Okay? In more general problems, in more general PTEs, this would just be called the forcing function. Okay? So this is the. Is the force and function in other pdes probably not other pdes. Maybe I should just call this in general pdes, all right, simply because the term body force may not have any significance in other pdes. All right. It's just, it's just the forcing function. Nevertheless, since we are talking in the context of, we are talking of elasticity let me state what this forcing function is. Okay. Give you, give you an example of the forcing function. So, as an example, let's suppose that we have again this sort of situation. We have the bar. Kay. It extends along the x direction. 0 and l, and that, I had this sort of picture that I drew early on, right? Just say that that is our distribution of the body force, right? F. So a specific example of this would be the following. Supposing that although I've drawn this body as extending in the horizontal direction at least relative to the, this particular screen let's also suppose that, interestingly enough in this situation, this was the direction of the action of the acceleration due to gravity. Okay? I just happened to turn my coordinate system around so that it turned out to be horizontal but that's the, the, the direction of gravity. Okay? So G is D. Acceleration- Due to- Gravity, okay? In that situation, f of x would be the mass density of the body, which, you know, could be a function of position, right? Maybe this bar does not have a constant density with respect to position times g. So is this a specific example of the body force. And know, perhaps make it a little more physical maybe we should say that this bar, you know, is actually turned around this way. So, downward is the x direction, and now you can see very clearly how that would be the body force, right? It could simply be the the distributed force. Distributed because it's distributed at every point in the, every, every, every, every little mass element of our bar. And it's the force that's distributed in every mass element of our bar due to gravity. Okay, so that's another example of it. Okay. So, I think we can stop this segment here. When we come back, we will continue with, saying a little more about this, differentiation.
